CBSE Class 12 Physics 2020 Delhi Set 3 Paper

Question : 3 of 7

Marks: +1, -0

SECTION - C

(a) Define the term 'half-life' of a radioactive substance.

(b) The half life of

{}_{92}^{238}\mathrm{U}

undergoing alpha decay is

4.5 \times 10^{9}

years calculate the activity of

5\ \mathrm{g}

sample of

{}_{92}^{238}\mathrm{U}

Solution:

(a) Half-life : The half-life of a radioactive substance is the time required for half the number of the nuclei to decay.

(b) Half-life

= 4.5 \times 10^{9}

years

\;\text{Mass number}\;\;=238

\lambda\;=\;\frac{0.693}{T}=\;\frac{0.693}{4.5 \times 10^{9}} \;\text{per year}\;

N\;=\;\frac{5 \times 6.02 \times 10^{23}}{238}

Hence, the activity

= A = \lambda N

\;\;\text{Or,}\;\; \;\; A=\;\frac{0.693}{4.5 \times 10^{9}} \times \;\frac{5 \times 6.02 \times 10^{23}}{238}

\;\therefore A=0.00195 \times 10^{15}

=1.95 \times 10^{12} \;\text{per year}\;

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